Driving transistors?

Discussion in 'General Electronics Chat' started by danielb33, Nov 4, 2012.

  1. danielb33

    Thread Starter Member

    Aug 20, 2012
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    I am familiar with driving FETS. Pushing a high voltage pulling a low to make my FET behave in a state according to my needs is easy. I have never had to use transistors. How do people make sure they are getting a specific current or zero current at 100KHz for a switching application? I do not need this now, but am curious about how this works. Controlling states in transistors seems more difficult than FETS. Any other comments are appreciated!

    Thanks.
     
  2. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    You know what FET stands for, right?

    Also, are you talking about the mode of operation: linear, cutoff, and saturation?
     
  3. Papabravo

    Expert

    Feb 24, 2006
    10,166
    1,797
    About the same really, except you don't have such long delays for charging and discharging the gate capacitance.

    1. When the transistor is in cutoff Vbe < 0.7 volts
    2. When the transistor is in satuaration Vbe ≈ 0.7V
    3. In the linear region Ic ≈ (hfe)(Ib)

    Drawing a load line on the collector curves to establish a Q-point is the same exercise as for a FET.

    In logic design there are D-FF designers and JK-FF designers. It is the same kind of thing for transistors.
     
  4. crutschow

    Expert

    Mar 14, 2008
    13,053
    3,244
    If you use them as switches a bipolar junction transistor (BJT) basically has a current operated input with a base-emitter drop of about 0.7V when on. You just need to apply a current to the base of about 1/10 of the expected collector current to fully turn them on. This current is typically established by a voltage applied to a resistor in series with the base. With no base current they are off.

    One disadvantage of BJT's as switches is that they can have a significant storage-time delay when turned off which FET's don't have.

    A field-effect transistor (FET) has a voltage operated input, as you know.
     
    Last edited: Nov 8, 2012
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